Python 如何设置SciPy插值器以最准确地保存数据

在我看来，您混淆了由interp1d构造的插补器对象和实际的插补坐标，后者是您想要的最终结果.

似乎在默认情况下，SciPy以均匀的间隔进行内插

interp1d将返回根据您提供的x和y坐标构建的插补器对象.它们根本不需要均匀分布.

然后，您向该模块提供xnew个值，这些值定义了该模块将在何处重建信号.这是你必须指定是否要均匀间隔的地方:results = interpolator(np.linspace(0, 1, num_interpolation_points)).T.tolist().注意对np.linspace的调用，它的字面意思是"线性间隔的值".

将其替换为np.logspace()，以获得几何间隔值，或者替换为其他值:

import numpy as np
from scipy.interpolate import interp1d

import matplotlib.pyplot as plt

# Generate fake data
x = np.linspace(1, 3, 1000)
y = (x - 2)**3

# interpolation
interpolator = interp1d(x, y)

# different xnews
N = 20
xnew_linspace = np.linspace(x.min(), x.max(), N)  # linearly spaced
xnew_logspace = np.logspace(np.log10(x.min()), np.log10(x.max()), N)  # log spaced

# spacing based on curvature
gradient = np.gradient(y, x)
second_gradient = np.gradient(gradient, x)
curvature = np.abs(second_gradient) / (1 + gradient**2)**(3 / 2)
idx = np.round(np.linspace(0, len(curvature) - 1, N)).astype(int)
epsilon = 1e-1

a = (0.99 * x.max() - x.min()) / np.sum(1 / (curvature[idx] + epsilon))
xnew_curvature = np.insert(x.min() + np.cumsum(a / (curvature[idx] + epsilon)), 0, x.min())

fig, axarr = plt.subplots(2, 2, layout='constrained', sharex=True, sharey=True)

axarr[0, 0].plot(x, y)
for ax, xnew in zip(axarr.flatten()[1:], [xnew_linspace, xnew_logspace, xnew_curvature]):
    ax.plot(xnew, interpolator(xnew), '.--')

axarr[0, 0].set_title('base signal')
axarr[0, 1].set_title('linearly spaced')
axarr[1, 0].set_title('log spaced')
axarr[1, 1].set_title('curvature based spaced')

plt.savefig('test_interp1d.png', dpi=400)

请注意，我不确定像我所做的那样对曲率进行zoom 是正确的方法.但这给了你大约interp1d个的 idea .